How do you find all the zeros of #f(x) = x^3+ 8x^2- x - 8#?

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Answer 1 · 2016-03-22T02:53:45

#x=-8,-1,1#

Notice that the coefficients of each term, when added, equal #0#:

#1+8-1-8=0#

This means that #f(1)=0#, so #1# is a zero of the function and #x-1# is a factor.

Use polynomial long division or synthetic division to determine that the other two zeros can be found from:

#(x^3+8x^2-x-8)/(x-1)=x^2+9x+8#

To find the remaining zeros of

#x^2+9x+8=0#

Factor the quadratic.

#(x+1)(x+8)=0#

Thus, if either of these terms equals #0#,

#x+1=0" "=>" "x=-1#

#x+8=0" "=>" "x=-8#

So, the function's three zeros are located at #x=-8,-1,1#.

We can check a graph of the function:

graph{x^3+8x^2-x-8 [-12, 4, -33.4, 85.3]}